Yep. Tile in the following pattern: 1,1,0,1,1,0,...,1,1 2,2,1,2,2,1,...,2,2 2,2,1,2,2,1,...,2,2 1,1,0,1,1,0,...,1,1 2,2,1,2,2,1,...,2,2 2,2,1,2,2,1,...,2,2 ......................... 1,1,0,1,1,0,...,1,1 Note that the height is 1 mod 3, and the width is 2 mod 3, which fits 40x41 perfectly.

My construction. Put shown $6\times 6$ squares together to form $42\times 42$, now delete the first and second row and the last column to obtain $40 \times 41$ table. [asy][asy] size(6cm); for (int i=0; i<=5; ++i) { draw(shift(i,0)*unitsquare); } for (int i=0; i<=5; ++i) { draw(shift(i,1)*unitsquare); } for (int i=0; i<=5; ++i) { draw(shift(i,2)*unitsquare); } for (int i=0; i<=5; ++i) { draw(shift(i,3)*unitsquare); } for (int i=0; i<=5; ++i) { draw(shift(i,4)*unitsquare); } for (int i=0; i<=5; ++i) { draw(shift(i,5)*unitsquare); } label("$1$", (1/2,1/2)); label("$1$", (3/2,1/2)); label("$0$", (5/2,1/2)); label("$1$", (7/2,1/2)); label("$1$", (9/2,1/2)); label("$0$", (11/2,1/2)); label("$2$", (1/2,3/2)); label("$2$", (3/2,3/2)); label("$1$", (5/2,3/2)); label("$2$", (7/2,3/2)); label("$2$", (9/2,3/2)); label("$1$", (11/2,3/2)); label("$2$", (1/2,5/2)); label("$2$", (3/2,5/2)); label("$1$", (5/2,5/2)); label("$2$", (7/2,5/2)); label("$2$", (9/2,5/2)); label("$1$", (11/2,5/2)); label("$1$", (1/2,7/2)); label("$1$", (3/2,7/2)); label("$0$", (5/2,7/2)); label("$1$", (7/2,7/2)); label("$1$", (9/2,7/2)); label("$0$", (11/2,7/2)); label("$2$", (1/2,9/2)); label("$2$", (3/2,9/2)); label("$1$", (5/2,9/2)); label("$2$", (7/2,9/2)); label("$2$", (9/2,9/2)); label("$1$", (11/2,9/2)); label("$2$", (1/2,11/2)); label("$2$", (3/2,11/2)); label("$1$", (5/2,11/2)); label("$2$", (7/2,11/2)); label("$2$", (9/2,11/2)); label("$1$", (11/2,11/2)); [/asy][/asy]